Covering map problem The following is a qualifying exam problem and I have no idea where to start.  Any help is appreciated.  Thank you.
$S^2$ is the standard sphere of radius 1.
Let X denote the space $S^2\bigcup A$, where $A =\{(x, 0, 0) \in \mathbb{R}^3: 1 \leq x \leq 2\}$.  Show that if $p : X \rightarrow Y$ is a covering map, then $p$ must be a
homeomorphism, i.e. $X$ cannot cover anything except itself.
 A: I don't know your background but if you know the properties of map lifting then here is a proof.
Let $p_1:S^2\rightarrow S^2$ be the identity covering map and if possible let $p_2:X\rightarrow S^2$ be another covering. Now consider the identity map between the base space and denote it by $i$. The $i\circ p_1$ will give you a map from $S^2\rightarrow S^2$ which can be lifted (as $\pi_1(S^2)=e$) to get a map $G=\tilde{i\circ p_1}:S^2\rightarrow X$. Similarly lift $i^{-1}\circ p_2=i\circ p_2$ to get a map $H=\tilde{i\circ p_2}:\rightarrow S^2$. $G$ and $H$ are continuous by construction. Show that they are inverse of each other. This will prove that $S^2$ and $X$ are homeomorphic.  
A: First, observe that $X$ is "nice" (=locally euclidean) in every point but one: it is the union of a sphere with a "tail" segment. Call $p$ the point of junction between the sphere and the segment, i.e. $p=(1,0,0)$.
Let $f:X \to Y$ a covering map. If its degree (= cardinality of the fiber over each point) is $1$, then this is just the trivial covering $X \to X$. So let's assume the degree is at least $2$. Let $z$ be the image of your singular point $p$ under this covering projection. Since the degree is at least $2$, you have that $f^{-1}(z)$ contains at least $2$ points, i.e. $p$ and at least one more. Now recall that a covering map is a local homeomorphism, so $z$, $p$ and any other point in $f^{-1}(z)$ are forced to have a homeomorphic neighborhood. But in $X$ there is no other point with the same local topology of $p$: every other point has local neighborhoods which are homeomorphic to open intervals or open balls; $p$ is the only point of $X$ all neighborhoods of which are of the form "segment+ball". So we have reached an absurd.
